As it is difficult to predict order using data provided in graph.
For specific time interval $0-5 \mathrm{sec}, 5-10 \mathrm{sec}$ and $10^{-15} \mathrm{sec}$. order comes to be zero, but graph is not a straight line.
Assuming $1^{\text {st }}$ order kinetics
$K=\frac{1}{t} \ln \frac{A_{0}}{A_{t}}$
$K=\frac{1}{10} \ln \frac{40}{20}$
Time required to reduce to $2.5 g / L$
$K=\frac{1}{t} \ln \frac{50}{2.5}$
$\frac{1}{10} \ln 2=\frac{1}{t} \ln 20$
$t=\frac{1.3010 \times 10}{0.3010}=43.3 \mathrm{sec}$.